a long rod ABC of mass m and length l has two particles of masses m and 2m attached to it as shown in the fig. the system is initially in the horizontal position the work done it keep it verticalily with A is at the bottom is
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a long rod ABC of mass m and length l has two particles of masses m and 2m attached to it as shown in the fig. the system is initially in the horizontal position the work done it keep it verticalily with A is at the bottom is
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I have attached the diagrams of before and after the rotation condition
Work done = change in potential energy
Assuming the horizontal condition to be a reference and potential energy is condidered zero at the reference.
So change in potential energy of the
2m mass
= 2mgl
m mass
=mgl/2
and for the rod lets assume that whole mass is centered at centere of mass at l/2
so change in potential energy
=mgl/2
net change in potential energy
= 2mgl+mgl/2+mgl/2
=3mgl